 This lecture is part of an online algebraic geometry course on schemes and will be about ample and very ample sheaves. So I'm going to simplify things a bit and we're going to take X to be a projective variety over an algebraically closed field K. Now you can define ample and very ample sheaves and much greater generality, but the definitions become somewhat more complicated. So I'm going to keep things as simple as possible. So let's take else be an invertible sheaf over X. And we saw last lecture that if you've got some sections. S naught up to SN of L over X. These give us a morphism from some open subset you contained an X to some projective space. And I mentioned one thing we could do is let's just take S naught to SN to be a basis for the global sections of of L. Which is finite dimensional because we're assuming X is a projective variety. And we can ask does this give an isomorphism from X to its image in P to the N. And if so, L is called very ample. So and we will say L is ample. If L to the N is very ample, some N greater than zero. I should warn you that these are kind of slightly cheap definitions that were okay for projective varieties, but these definitions are not okay for general schemes. If you want to see the definition of general skip for general schemes, you've got a bit of a problem because there seem to be several slightly different and in equivalent definitions of these four general schemes. So you've checked the exact details rather carefully, but you can find things like for instance is obvious with this definition that very ample sheaves are ample. For some of the definitions for general schemes is not always true that very ample sheaves are ample so you really have to be a little bit careful if you're not doing projective varieties. So both of these conditions are sort of positivity conditions on sheaves. So there are several positivity conditions you can put on sheaves like saying they're ample or very ample or generated by sections and other things like neph and so on. And positive sheaves tend to have lots of global sections if they're sufficiently positive. And if sheaves are sufficiently positive they also tend to have vanishing higher cohomology groups. So ample and very ample are two notions of positivity. So let's have some examples. Let's look at the sheaf O N on projective space. Now for N less than zero, this is no global sections other than zero. For N equals zero the only global sections are constants. So for any zero the only sections are constant. So what we do is we get a map from P to the K to a point, which is certainly not an isomorphism to its image so this isn't ample either however for N equals one. If we look at all the global sections of O one that they're essentially X naught up to X N. And we just get the an isomorphism from P K to P K. So so this is very ample for N greater than one. Again, we get an isomorphism to its image so for P one. We looked at this last time and we've got things like maps from P one to P one or a quadric. But N equals two or the the twisted cubic if any equals three and so on. So we see that O N is ample. N is greater than or equal to one and you could put in the word very there if you want in this case very ample sheaves are the same as ample sheaves. So now we should give an example of a line bundle that is ample but not very ample and for this we're going to take X to be an elliptic curve. And we're going to cheat a bit and work over the complex numbers and working over complex numbers makes life very easy because we can just use elliptic functions and be frowned upon by algebraic geometers who think we should do everything algebraically. So we're just going to take X to be the elliptic curve over L where I guess I shouldn't use L for a line bundle and a lattice. So what the heck here L is just going to be a lattice just to have to get used to me using L for two different things. And we're going to take the line bundle L is that is the bun is is the sheaf of the divisor zero so you remember we can form a divisor on X just by taking a finite number of points with multiplicities. And we're just going to take the point zero with multiplicity one. And then we can ask what are the sections of L to the N. Well, for N less than zero. There are going to be no sections. So I should have said the sections, these just correspond to to elliptic functions with a pole of order less than we would have N at zero and holomorphic elsewhere. So if N is zero, then the only sections of L to the N is a zero section for N equals zero. The only section is is is a constant for N equals one. You remember there are no elliptic functions with just a single pole of order one so apart from constant. So again just get one for N equals two, we get the vice stress P function, and we also get one because this has a double pole at the origin for N equals three. We get the vice stress function we also get its derivative, which has a pole of order three, and we again get one for N equals four, we get the vice stress function with a double pole. We get its derivative, we get it square, and we get one. So let's see what happens with these well for these two. We only get a map from the elliptic curve to a point. So these are certainly not very ample for N equals three. We get an isomorphism from L to the elliptic curve, y squared equals for x cubed minus g to x minus g three by mapping where we take y equals the vice stress function at a point Z. So that should be the derivative and x is the vice stress function at point Z. So this is the famous differential equation satisfied by the vice stress elliptic function. So for any equals three we get something that is very ample for any equals two. It's not very ample because here we're getting a map from the elliptic curve to P one. And this is just mapping a point x, y, z to the point x, y. I've got that right. Not to the point x, y, to the point x, one. So we're just essentially mapping the elliptic curve to the x axis, which is usually a two to one mapping. And it's not an isomorphism to its image. So this is not very ample. However, these two things here are at least ample. So the very ample line bundles are anything with any power of O, N with any power of L, any Nth power of L with N at least three, whereas the ample ones are these ones here. So L and L squared are ample but not very ample and L cubed and above are all very ample. Next, we should look at the following problem. Given a line bundle L on a projective variety X and sections S naught SN of L, when do they define an isomorphism from X to its image in P to the N? In other words, when is this, when do they define a closed immersion? So let's have some necessary conditions. First of all, the function, the morphism must be defined everywhere. Must be defined on the whole of X. You remember, in general, it's only defined on an open subset of X. And for it to be defined on the whole of X, S naught up to X SN must generate L. You remember this means that at each point of X, their restrictions that point must generate the stork of L at that point. Secondly, the morphism to PN is injective. And what this means is that there are enough sections that you can distinguish two points on X. So given X, Y in X, we can find the section in the linear combination of S naught up to SN, which is say zero on X, none zero on Y. So you can say whether a section of an invertible sheaf is zero or not, because you just have to look at that point and see whether its image is in the maximal sub module of the stork or not. So for this, we say that the sections S I separate points. And if they separate points, we get an injective map from extra projective space. We can ask, is this enough? And the answer is no. If we have these two conditions satisfied, it doesn't necessarily mean that we've got a closed immersion from X to its image. For example, let's look at the following map from P1 to P2. What we do is we take X colon Y to let's take it to X cubed, X squared, Y, Y cubed. And you can check this is injective and defined everywhere. However, if we look at the image, it sort of looks like this. The image, if we just look at the points where Y equals one and look at X cubed, X squared, then we see what we really, the image of P1 kind of has this funny singularity. We've got a curve with a cusp. So this map is injective, but it's not an isomorphism onto its image. And let's have a look at why not. If you've got a P1 here mapping to it there, if we look at this point here, we can see that a tangent vector at this point gets mapped to zero. So it's not injective on tangent spaces. Obviously, it has to be injective on tangent spaces if this is a closed immersion. So we need the extra condition that the embedding should be injective on tangent spaces. And what this corresponds to is the sections of L at X vanishing at X have to span the Mx Lx over Mx squared Lx, where this is going to be the stalk of L at X. And this is the maximal ideal of the local ring at X. So what this is essentially saying that the sections of L kind of span the cotangent space. And that's enough to sort of show that they separate tangent vectors. So we have this condition of saying that the sections separate tangent vectors. And so this is condition three that the sections must generate the line bundle. They must separate points and they must separate tangent vectors. So separating points and tangent vectors is a zero thought on a first order condition. And you might think we also need some sort of second order condition and a third order condition and so on, but fortunately we don't. This is enough to show that the induced map from X to P to the N is a closed immersion. And let's see why. Well, to show it's a closed immersion. We essentially just need to show the map from O from the chief of regular functions on projective space to F star of O of X is surjective. This is because a closed immersion is something that looks locally like a map from spectrum of A. So a map from the spectrum of B, the spectrum of A where the map from rings A to B is on to. So we can check this at Storks. So what we want to do is to show that we want to show that at each point the map from A to B is on to where this is the local ring. Of P to the N at the image of some point. And this is the local ring of F star O of X at the same point F of X. So we've just got to check that this map of local rings is on to for each point and that will be enough to show we've got a closed immersion. So let's assemble the information we've got about A and B in order to show that on to. So summary is we've got a map F from A to B of local rings. It's a local homomorphism. And what do we know about it? Well, A over M A is isomorphic to B over M B where here M A and M B are the maximal ideals. And this is because these are both the same as the field K we're working over. Secondly, we know the map from M A over M A squared to M B over M B squared is on to. So this is just the condition about separating tangent vectors. Thirdly, we know that B is a finitely generated a module. This follows because F star of a coherent chief such as O of X is coherent, which follows because F is projective and projective maps map coherent sheaves to coherent sheaves as we did earlier. So we want to stop from this information and so we want to show that F is on to. And this is quite easy to do. What we do is we recall Nakayama's lemma, which says that if N is a finitely generated module over R, which is local, then N equals M N implies N equals zero. Here M is, of course, the maximum ideal of R. So now let's prove this. First of all, we take a B module M which is equal to M B over M A times B. And what we want to do is to show that M A times B is equal to M B. So we want to show that M equals zero. Well, we observe that M over M B times M is equal to zero. And this follows from this fact here. So M equals zero. And now we're applying Nakayama's lemma. So M B equals M A dot B because M is zero. Now, instead of looking at a B module, we look at the A module B over F of A. And let's call this N. And then we notice that N over M A times N is equal to B over M A times B. And we want that to be generated by F A, which is equal to B over M B times and F A. And here we're using this fact here. And now this is equal to naught. And now we're using Nakayama a second time. So B over F A equals naught. So B equals F of A, which is what we wanted to prove. Prove that a map from F from A to B is surjective. You can combine this result with the Riemann-Roch theorem. If X is a projective curve and L is an invertible sheaf, then L is ample. It's the same as saying that the degree of L is greater than zero. Here we remember that invertible sheafs can be identified with divisors, which is of the form sum of Ni Pi, where the Pi are points on X and the Ni are integers. And the degree is just sum of Ni. So this gives a very simple criterion for curves, for invertible sheafs on curves to be ample. And what we do is we use the result. We've just proved that a sheaf is ample, if and only if it's got plenty of sections satisfying various conditions. The Riemann-Roch theorem that comes later shows that if a divisor has sufficiently high degree, then it has enough sections satisfying the conditions we've just mentioned. So that the invertible sheaf, any invertible sheaf of degree greater than zero is ample. Okay, next lecture will be more about the relation between invertible sheaves and projective embeddings.